The percentage purity of commercial #"HCl"# is 35% (w/w) & the specific gravity of the solution is 1.18 #"g"//"mL"#. What are the molarity, molality and normality?

1 Answer
Truong-Son N.
Jul 17, 2015

I happen to know (from using the following) that #37%# reaction-grade #"HCl"# is about #"12.06 M"#... so let's go off of that. I also looked that up here. You can also go off of #36%# #"HCl"# which is about #"11.65 M"# (easily googlable).

It's dissolved in water, and #"HCl"# is

#35.453+1.0079 = "36.4609 g/mol"#.

#"1 L"# of the solution is

#"1.18 g"/cancel"mL"cdot (1000cancel("mL"))/cancel("1 L")cdotcancel"1 L" = "1180 g"#.

Thus, #35"% w/w"# for #"HCl"# puts it at #"413 g HCl"/"1180 g soln"#.

Molarity is #("mol solute")/("L solution")#, thus it is:

#"Molarity" = [413cancel("g")*("1 mol")/(36.4609cancel("g"))]/("1 L") ~~ color(blue)("11.3272 M")#

which is sensible considering how #37%# is close to #"12 M"#.

Molality is #("mol solute")/("kg solvent")#.

The density of water at #25^@"C"# is #"0.9970749 g/mL"#. Thus, with #"1 L"# of water, it has a mass of #"0.9970749 kg"#. Then:

#"Molality" = [413cancel("g")*("1 mol")/(36.4609cancel("g"))]/("0.9970749 kg") ~~ color(blue)("11.3604 m")#

Since #HCl# is monoprotic, its normality is equal to its molarity. Thus, its normality is also:

#"Normality"# #~~# #color(blue)("11.3272 N")#

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